DS2011

c = a + b

a = 2b - 15

a + b + c = 180

So, 2a + 2b = 180

Plugging in 2(2b - 15) for 2a, we get:

4b - 30 + 2b = 180

6b - 30 = 180

6b = 210

b = 35

a = 2(35) - 15

a = 70 - 15

a = 55

55 + 35 = c

c = 90

a = 55, b = 35, c = 90

b = 35

Answer: 35

By the Exterior Triangle Angle Sum Theorem ( kinda forgot name) , angle ABC + angle BAC = angle ACD, so

\(x^2-4x-3=9x+27\)

\(x^2-13x-30=0\)

I will use the quadratic formula to find the roots of x, and then eliminate the value of x that produces negative angle measures.

a = 1, b = -13, c = -30

\(x = {13 \pm \sqrt{169+120} \over 2}\)

\(x = {13 \pm 17 \over 2}\)

We get: x = 15, and x = -2.

Eliminate x = -2, and the answer is x = 15

Answer: 15

Let's simplify:

\(\frac{a}{3}+1=1+\frac{3}{a}-a+\frac{2}{a}\)

\(a-\frac{5}{a}+\frac{a}{3}=0\)

I just put this on a calculator and got a = \(\frac{\sqrt 15}{2}\), \(\frac{-\sqrt 15}{2}\)

I didn't know what to do.

Answer: (sqrt 15)/2, -(sqrt 15)/2

First time, in mph was 2.5 x 3 = 7.5 mph. 

Last time, she ran 5.2 - 3.8 = 1.4 miles, in mph was 1.4 x 3 = 4.2 mph.

7.5 - 4.2 = 3.3 miles.

Answer: 3.3 miles

\(f(x) = \frac{(2x + 3)(x - 7)}{\sqrt{x}}\)

The domain must be all real numbers except for when \(\sqrt x\) = 0.

x can not equal 0, so the answer is (-inf, 0) U (0, inf)

Answer: (-inf, 0) U (0, inf)

We can see that this olygon has a rotational symmetry of 120 degrees, since rotating the figure 120 degrees makes the same shape.

360 degrees/120 degrees = 3 rotations

Answer: 3

I think to do n(n+1)(n+2)(n+3)=461^2 -1, then do

(n^2 +n)(n+2)(n+3)=212520

(n^3+3n^2+2n)(n+3)=212520

n^4+3n^3+3n^3+9n^2+2n^2+6n=212520

n^4 + 6n^3 + 11n^2 + 6n = 212520

What now? Or is there another way to do this?

Pls help

A(2) = 3 - 40 x  4 x 4 = 3 - 640 = -637.

A(1) = 3 - 5 x 4 x 2 = 3 - 40 = -37

-637 - ( - 37 ) = -637 + 37 = -600

Answer: -600

Answer= D

Answer: B